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A cross-flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine (a) the exit temperature of the hot fluid and \((b)\) the rate of heat transfer in the heat exchanger.

Short Answer

Expert verified
Answer: The exit temperature of the hot fluid is approximately \(27^{\circ} \mathrm{C}\), and the rate of heat transfer in the heat exchanger is approximately \(2,280,800 \mathrm{~W}\).

Step by step solution

01

Identify the given parameters

We are given the following parameters: - Overall heat transfer coefficient (U): \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Heat transfer surface area (A): \(400 \mathrm{~m}^{2}\) - Heat capacity of hot fluid (Cp_h): \(40,000 \mathrm{~W} / \mathrm{K}\) - Heat capacity of cold fluid (Cp_c): \(80,000 \mathrm{~W} / \mathrm{K}\) - Inlet temperature of hot fluid (T1_h): \(80^{\circ} \mathrm{C}\) - Inlet temperature of the cold fluid (T1_c): \(20^{\circ} \mathrm{C}\)
02

Determine the minimum heat capacity rate and the heat capacity rate ratio

First, we need to find the minimum heat capacity rate, which is the smaller of Cp_h and Cp_c: \(C_{min} = min(Cp_h, Cp_c) = min(40,000, 80,000) = 40,000 \mathrm{~W} / \mathrm{K}\) Next, calculate the heat capacity rate ratio (C_r): \(C_r = \dfrac{C_{min}}{C_{max}} = \dfrac{40,000}{80,000} = 0.5 \)
03

Calculate the heat transfer effectiveness of the heat exchanger (ε)

The heat transfer effectiveness (ε) of a cross-flow heat exchanger with both fluids unmixed can be found using the following formula: \(\varepsilon = 1 - \mathrm{exp} \left(- Ntu \cdot (1 + C_r) \right)\) Where Ntu is the number of transfer units and is calculated as: \(Ntu = \dfrac{UA}{C_{min}}\) Calculate Ntu: \(Ntu = \dfrac{200 \cdot 400}{40,000} = 2\) Now, substitute the Ntu and C_r values back into the formula for ε: \(\varepsilon = 1 - exp(-2 \cdot (1 + 0.5)) = 1 - exp(-3) \approx 0.9502\)
04

Calculate the actual heat transfer rate (Q)

Now that we have the heat transfer effectiveness (ε), we can calculate the actual heat transfer rate (Q). The formula for Q is: \(Q = \varepsilon \cdot C_{min} \cdot (T1_h - T1_c)\) Substitute the given values and the calculated ε into the formula: \(Q = 0.9502 \cdot 40,000 \cdot (80 - 20) \approx 2,280,800 \mathrm{~W}\)
05

Determine the exit temperature of the hot fluid (T2_h)

Finally, we can find the exit temperature of the hot fluid (T2_h) using the equation: \(T2_h = T1_h - \dfrac{Q}{Cp_h}\) Substitute the given values and the calculated Q into the equation: \(T2_h = 80 - \dfrac{2,280,800}{40,000} \approx 27^{\circ} \mathrm{C}\) Thus, (a) the exit temperature of the hot fluid is approximately \(27^{\circ} \mathrm{C}\) (b) the rate of heat transfer in the heat exchanger is approximately \(2,280,800 \mathrm{~W}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Overall Heat Transfer Coefficient
When dealing with heat exchangers, one of the fundamental concepts is the overall heat transfer coefficient, often denoted by \(U\). This coefficient represents how well heat is transferred between two fluids separated by a solid wall. It encompasses both conduction through the wall and convection away from the surface.

The formula used is:
  • \(U = \frac{Q}{A \Delta T_{lm}}\)
where \(Q\) is the rate of heat transfer, \(A\) is the heat transfer area, and \(\Delta T_{lm}\) is the log mean temperature difference.
Factors influencing \(U\) include:
  • Material properties: Conductivity of the wall materials.
  • Flow conditions: Speed and type of fluid movement.
  • Temperature differences: The warmer the fluid, the greater the potential for heat flow.
Understanding \(U\) helps in designing efficient systems that optimize energy usage.
Heat Capacity Rate Ratio
The heat capacity rate ratio, abbreviated as \(C_r\), is a crucial factor in heat exchanger design and performance analysis. It is defined as the ratio of the minimum to the maximum heat capacity between the hot and cold fluids.

Mathematically, it is expressed as:
  • \(C_r = \frac{C_{min}}{C_{max}}\)
where \(C_{min}\) and \(C_{max}\) are the smaller and larger heat capacities, respectively.
A few points to consider:
  • Influence on effectiveness: \(C_r\) affects the effectiveness of the heat exchanger, with lower values generally leading to higher effectiveness.
  • Symmetry: When the heat capacities are equal, \(C_r\) equals 1, simplifying analysis.
By knowing \(C_r\), engineers can predict heat exchanger performance and achieve the desired thermal outputs.
Effectiveness-NTU Method
The Effectiveness-NTU method is a widely used approach for analyzing heat exchanger performance. NTU stands for 'Number of Transfer Units,' which quantifies the thermal size of the exchanger relative to the heat capacity rate.

Key formulas include:
  • \(Ntu = \frac{UA}{C_{min}}\)
  • \(\varepsilon = 1 - \exp(- Ntu \cdot (1 + C_r))\)
Where \(\varepsilon\) is the effectiveness, indicating how well the heat exchanger transfers heat compared to the maximum possible.Benefits of this method:
  • Simplicity: Directly relates heat transfer to physical attributes of the exchanger.
  • Flexibility: Can be adapted to various configurations and flow patterns.
Understanding the Effectiveness-NTU method allows engineers to solve complicated thermal problems with straightforward calculations.
Cross-Flow Exchangers
Cross-flow heat exchangers are a common configuration where two fluids flow perpendicular to each other. This setup is prevalent because it maximizes surface area and takes advantage of space.

Details of cross-flow exchangers:
  • Fluid arrangement: In 'unmixed' configurations, fluid streams remain isolated, enhancing temperature change.
  • Applications: Used in systems like air conditioning and radiators.
Analyzing cross-flow exchangers involves:
  • Calculating effectiveness based on flow arrangement using specific equations for 'mixed' or 'unmixed' flow.
By understanding this configuration, engineers can tailor heat exchanger designs to specific thermal requirements, allowing for efficient heat management in diverse applications.

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Most popular questions from this chapter

The cardiovascular counter-current heat exchanger has an overall heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Arterial blood enters at \(37^{\circ} \mathrm{C}\) and exits at \(27^{\circ} \mathrm{C}\). Venous blood enters at \(25^{\circ} \mathrm{C}\) and exits at \(34^{\circ} \mathrm{C}\). Determine the mass flow rates of the arterial blood and venous blood in \(\mathrm{g} / \mathrm{s}\) if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and the surface area of the heat transfer to occur is \(0.15 \mathrm{~cm}^{2}\).

11-100 E(S) Reconsider Prob. 11-99. Using EES (or other) software, investigate the effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and the surface area. Let the inlet temperature vary from \(60^{\circ} \mathrm{C}\) to \(120^{\circ} \mathrm{C}\) and the overall heat transfer coefficient from \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) to \(1250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Plot the rate of heat transfer and surface area as functions of the inlet temperature and the heat transfer coefficient, and discuss the results. 11-101E A thin-walled double-pipe, counter-flow heat exchanger is to be used to cool oil \(\left(c_{p}=0.525 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(300^{\circ} \mathrm{F}\) to \(105^{\circ} \mathrm{F}\) at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) by water \(\left(c_{p}=\right.\) \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\) ) that enters at \(70^{\circ} \mathrm{F}\) at a rate of \(3 \mathrm{lbm} / \mathrm{s}\). The diameter of the tube is 5 in and its length is \(200 \mathrm{ft}\). Determine the overall heat transfer coefficient of this heat exchanger using (a) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hotand cold-water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same.

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the \(\varepsilon-\mathrm{NTU}\) method.

A shell-and-tube heat exchanger with 2-shell passes and 8 -tube passes is used to heat ethyl alcohol \(\left(c_{p}=2670 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(25^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(2.1 \mathrm{~kg} / \mathrm{s}\). The heating is to be done by water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(95^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger.

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